If $\alpha$ is of bounded variation on $[a,b]$, then it is continuous almost everywhere on $[a,b]$ If $\alpha$ is of bounded variation on $[a,b]$, then it is continuous almost everywhere on $[a,b]$.
I know that a function is of bounded variation iff it is the difference of two monotone functions. Monotone functions have a countable number of jump discontinuities hence the original function is continuous almost everywhere. This makes sense intuitively, but how to make this rigorous?
 A: Any class that would reject the above proof is probably emphasizing formality over clarity.
Still, you can say more explicitly:


*

*If $f=g-h$ where $f,g,h$ are functions on $[a,b]$, and the set of discontinuities of $g$ is $U$ and the set of discontinuities of $h$ is $V$, then the set of discontinuities of $f$ is a subset of $U\cup V$.

*In particular, if $g,h$ are continuous almost everywhere, then $g-h$ is continuous almost everywhere. (The union of two measure zero sets is of measure zero. Any subset of a measure zero set is measure zero.)

*If $f$ is of bounded variation, then $f=g-h$ for some pair of monotonic functions, $g,h$.

*Monotonic functions have at most countably many discontinuities. Finite and countable sets are measure zero. So monotonic functions are continuous almost everywhere.

*Therefore $f=g-h$ is continuous almost everywhere by $2$.
That's less clear than your outline, IMHO. I do think (1) is worth stating explicitly. In particular, the point about the discontinuities being a subset of the union of discontinuities. For example, if $g=h$, then $f=g-h=0$ is continuous everywhere, no matter how many discontinuities there are in $g=h$.
A: Just to add that if a function $\alpha$ is of bounded variation over a finite interval $[a,b]$, then  any point of continuity of $\alpha$ is a point of continuity of the variation function defined as $V(x)=0$ if $x=a$ and $V(x)=V(\alpha;[a,x])$ for $a<x\leq b$, and viceversa.
Here we use the notation
$$V(\alpha;[a,b])=\sup_{\mathcal{P}}\sum^n_{j=1}|\alpha(x_j)-\alpha(x_{j-1})|$$
where $\mathcal{P}=\{a=x_0<\ldots<x_n=b$ is a partition of $[a,b]$.
I presume that the additive properties of the variation function are known, that is, for any $a<c<b$
$$V(\alpha;[a,b])=V(\alpha;[a,c])+V(\alpha;[c,b])$$

Since $|\alpha(x)-\alpha(y)|\leq V(\alpha;[x,y])$ for all $a\leq x<y\leq b$, it is clear that any point of continuity of $V$ is a point of continuity for $\alpha$.
Now, suppose $a<x_0<b$ and that $\alpha$ is continuous at $x_0$. Given $\varepsilon>0$ choose $\delta>0$ and partitions $\{a=t_0<\ldots<t_k=x_0\}$ and $\{x_0=s_0<\ldots <s_m=b\}$ such that
if $|x-x_0|<\delta$, then $|\alpha(x)-\alpha(x_0)|<\tfrac\varepsilon2$, and
\begin{align}
V(\alpha;[a,x_0])-\frac\varepsilon2 &<\sum^n_{j=1}|\alpha(t_j)-\alpha(t_{j-1})|\tag{1}\label{one}\\
V(\alpha;[x_0,b])-\frac\varepsilon2 &<\sum^m_{k=1}|\alpha(s_k)-\alpha(s_{k-1})|\tag{2}\label{two}
\end{align}
Since adding point to a partition $P$ only increases the sum $\sum_j|\Delta \alpha_j|$, we may assume without loss of generality that $|x_0-t_{n-1}|<\delta$ and $|s_1-x_0|<\delta$. From \eqref{one} and \eqref{two} we get that
\begin{align}
V(\alpha;[a,x_0])<V(\alpha;[a,t_{n-1}]+\epsilon\tag{1'}\label{onep}\\
V(\alpha;[x_0,b])<V(\alpha;[s_1,b])+\varepsilon\tag{2'}\label{twop}
\end{align}
\eqref{onep} means that for $0<x_0-x<\delta$,
$$V(x_0)-V(x)<\varepsilon.$$
\eqref{twop} means that for $0<x-x_0<\delta$,
$$\begin{align}
V(x)-V(x_0)&=V(\alpha;[a,x])-V(\alpha;[a,x_0])\\
&=\big(V(\alpha;[a,b])-V(\alpha;[x,b])\big)-\big(V(\alpha;[a,b])-V(\alpha;[x_0,b])\big)\\
&=V(\alpha;[x_0,b])-V(\alpha;[x;b])<\varepsilon
\end{align}
$$
This proves continuity of $V$ at $x_0$.
The cases where $x_0=a$ or $x_0=b$ are handled similarly.
